How about Likert-type item?

How about Likert-type item?

Probability

• An event $E$ occurring within a particular sample space $S$
• Event: Expected results
• Sample space: All possible outcomes
• Probability $P$ is a proportion of event divided by its sample space
• Or mathematically:

$$P(E=e)=\frac{E}{S}$$

• Suppose we have a fair coin and doing a flip 10 times, where H indicates the head and T indicates the tail

Probability

• An event $E$ occurring within a particular sample space $S$
• Event: Expected results
• Sample space: All possible outcomes
• Probability $P$ is a proportion of event divided by its sample space
• Or mathematically:

$$P(E=e)=\frac{E}{S}$$

• Suppose we have a fair coin and doing a flip 10 times, where H indicates the head and T indicates the tail
• Then, our sample space:

set.seed(1)
S <- sample(c("H", "T"), 10, replace=TRUE, prob=rep(1/2, 2)) %T>% print()

##  [1] "T" "T" "H" "H" "T" "H" "H" "H" "H" "T"

Probability

• An event $E$ occurring within a particular sample space $S$
• Event: Expected results
• Sample space: All possible outcomes
• Probability $P$ is a proportion of event divided by its sample space
• Or mathematically:

$$P(E=e)=\frac{E}{S}$$

• Let the head be our expected outcome

Probability

• An event $E$ occurring within a particular sample space $S$
• Event: Expected results
• Sample space: All possible outcomes
• Probability $P$ is a proportion of event divided by its sample space
• Or mathematically:

$$P(E=e)=\frac{E}{S}$$

• Let the head be our expected outcome
• Then, our event:

E <- S[which(S == "H")] %T>% print()

## [1] "H" "H" "H" "H" "H" "H"

Probability

• An event $E$ occurring within a particular sample space $S$
• Event: Expected results
• Sample space: All possible outcomes
• Probability $P$ is a proportion of event divided by its sample space
• Or mathematically:

$$P(E=e)=\frac{E}{S}$$

• Thus, we can regard the probability of having a desired outcome as a relative frequency of events in a given sample space

Probability

• An event $E$ occurring within a particular sample space $S$
• Event: Expected results
• Sample space: All possible outcomes
• Probability $P$ is a proportion of event divided by its sample space
• Or mathematically:

$$P(E=e)=\frac{E}{S}$$

• Thus, we can regard the probability of having a desired outcome as a relative frequency of events in a given sample space
• As such:

length(E) / length(S)

## [1] 0.6

Determine the Probability

Determine the Probability

Caveats in enumeration

• Higher sample space $\to$ harder to solve
• It is more apparent with sequential problem
• Sequential problem: when you need to calculate probability from two different instances
• Example: the probability of having three 4 while rolling a dice three times

Determine the Probability

Caveats in enumeration

• Higher sample space $\to$ harder to solve
• It is more apparent with sequential problem
• Sequential problem: when you need to calculate probability from two different instances
• Example: the probability of having three 4 while rolling a dice three times

Tree diagram is available to solve a more complex probability problem

Determine the Probability

Sample case $\to$ the urn problem

• We have an urn filled with 30 blue and 50 red balls
• All balls are identical except for color
• In the urn, all balls have an equal distribution
• Task: Take three balls without replacement
• Question: How high is the chance of getting three blue balls?

Determine the Probability

    B (30/80)
   /
  /
80
  \
   \          
    R (50/80)

Determine the Probability

               B (29/79)
             /
    B (30/80)
   /         \
  /            R (50/79)
80
  \
   \          
    R (50/80)

Determine the Probability

                        / B (28/78)
               B (29/79)
             /          \ R (50/78)
    B (30/80)
   /         \
  /            R (50/79)
80
  \
   \          
    R (50/80)

Determine the Probability

                        / B (28/78)
               B (29/79)
             /          \ R (50/78)
    B (30/80)
   /         \
  /            R (50/79)
80
  \
   \          
    R (50/80)

The chance for having three blue balls is 0.0494

Determine the Probability

                        / B (28/78)
               B (29/79)
             /          \ R (50/78)
    B (30/80)
   /         \          / B (?)
  /            R (50/79)
80                      \ R (?)
  \
   \          
    R (50/80)

We have learnt how to draw a tree diagram. Now, what should we fill the question mark with?

Let's roll the dice :)

dice <- function(n) {
    sample(1:6, n, replace=TRUE, prob=rep(1/6, 6))
}

Let's roll the dice :)

dice <- function(n) {
    sample(1:6, n, replace=TRUE, prob=rep(1/6, 6))
}

Let's see whether our function work...

Let's roll the dice :)

dice <- function(n) {
    sample(1:6, n, replace=TRUE, prob=rep(1/6, 6))
}

Let's see whether our function work...

dice(1)

## [1] 3

It does!

Let's roll the dice :)

• So we shall roll the dice 10 times
• Let 4 be our outcome of interest
• How high is the probability of having the event within 10 trials?

Let's roll the dice :)

• So we shall roll the dice 10 times
• Let 4 be our outcome of interest
• How high is the probability of having the event within 10 trials?

set.seed(1)
roll <- dice(10) %T>% print()

##  [1] 3 4 5 1 3 1 1 5 5 2

• How high is the probability of getting 4?

Let's roll the dice :)

• So we shall roll the dice 10 times
• Let 4 be our outcome of interest
• How high is the probability of having the event within 10 trials?

set.seed(1)
roll <- dice(10) %T>% print()

##  [1] 3 4 5 1 3 1 1 5 5 2

• How high is the probability of getting 4?
• Turns out, it is 1/10

Let's roll the dice :)

• So we shall roll the dice 10 times
• Let 4 be our outcome of interest
• How high is the probability of having the event within 10 trials?

set.seed(1)
roll <- dice(10) %T>% print()

##  [1] 3 4 5 1 3 1 1 5 5 2

• How high is the probability of getting 4?
• Turns out, it is 1/10
• We have a fair dice, why is the probability not 1/6?

Let's roll the dice :)

• So we shall roll the dice 10 times
• Let 4 be our outcome of interest
• How high is the probability of having the event within 10 trials?

set.seed(1)
roll <- dice(10) %T>% print()

##  [1] 3 4 5 1 3 1 1 5 5 2

• How high is the probability of getting 4?
• Turns out, it is 1/10
• We have a fair dice, why is the probability not 1/6?
• Hint: sample and population

Let's roll the dice :)

• So we shall roll the dice 10 times
• Let 4 be our outcome of interest
• How high is the probability of having the event within 10 trials?

set.seed(1)
roll <- dice(10) %T>% print()

##  [1] 3 4 5 1 3 1 1 5 5 2

• How high is the probability of getting 4?
• Turns out, it is 1/10
• We have a fair dice, why is the probability not 1/6?
• Hint: sample and population
• The more sample we got, the closer it is to represent the population

Let's roll the dice :)

• What will we get with different number of rolls?

Let's roll the dice :)

• What will we get with different number of rolls?
• 100 rolls:

set.seed(1); roll <- dice(100)
sum(roll==4) / length(roll)

## [1] 0.25

Let's roll the dice :)

• What will we get with different number of rolls?
• 100 rolls:

set.seed(1); roll <- dice(100)
sum(roll==4) / length(roll)

## [1] 0.25

• 1,000 rolls:

set.seed(1); roll <- dice(1000)
sum(roll==4) / length(roll)

## [1] 0.2

Let's roll the dice :)

• What will we get with different number of rolls?
• 100 rolls:

set.seed(1); roll <- dice(100)
sum(roll==4) / length(roll)

## [1] 0.25

• 1,000 rolls:

set.seed(1); roll <- dice(1000)
sum(roll==4) / length(roll)

## [1] 0.2

• 10,000 rolls:

set.seed(1); roll <- dice(10000)
sum(roll==4) / length(roll)

## [1] 0.1724

Let's roll the dice :)

• 100,000 rolls:

set.seed(1); roll <- dice(100000)
sum(roll==4) / length(roll)

## [1] 0.1661

Let's roll the dice :)

• 100,000 rolls:

set.seed(1); roll <- dice(100000)
sum(roll==4) / length(roll)

## [1] 0.1661

• 1,000,000 rolls:

set.seed(1); roll <- dice(1000000)
sum(roll==4) / length(roll)

## [1] 0.1664

Let's roll the dice :)

• 100,000 rolls:

set.seed(1); roll <- dice(100000)
sum(roll==4) / length(roll)

## [1] 0.1661

• 1,000,000 rolls:

set.seed(1); roll <- dice(1000000)
sum(roll==4) / length(roll)

## [1] 0.1664

• 10,000,000 rolls:

set.seed(1); roll <- dice(10000000)
sum(roll==4) / length(roll)

## [1] 0.1666

Let's roll the dice :)

Let's roll the dice :)

Let's roll the dice :)

How high is the error in our trials?

Let's roll the dice :)

How high is the error in our trials?

• First we need to set the function to calculate error

epsilon <- function(p.hat, n) {
    sqrt({p.hat * (1-p.hat)}/n)
}

Let's roll the dice :)

How high is the error in our trials?

• First we need to set the function to calculate error

epsilon <- function(p.hat, n) {
    sqrt({p.hat * (1-p.hat)}/n)
}

• Get the roll and probability

roll <- c(10, 100, 1000, 10000, 100000, 1000000, 10000000)
prob <- sapply(roll, function(n) {
    set.seed(1); roll <- dice(n)
    sum(roll==4) / length(roll)
})

Let's roll the dice :)

df <- data.frame(list("roll"=roll, "prob"=prob))
df %>% knitr::kable() %>% kable_styling()

Let's roll the dice :)

df <- data.frame(list("roll"=roll, "prob"=prob))
df %>% knitr::kable() %>% kable_styling()

df$error <- mapply(function(p.hat, n) {
    epsilon(p.hat, n)
}, p.hat=df$prob, n=df$roll) %T>% print()

## [1] 0.0948683 0.0433013 0.0126491 0.0037773 0.0011769 0.0003724 0.0001178

Let's roll the dice :)

Here is a nice figure to summarize the concept:

Homework

Random Variables

Random Variables

• All sampled random variables should be independent from one another
• Each sampling procedure have to be identical, as to produce similar probability

Random Variables

• All sampled random variables should be independent from one another
• Each sampling procedure have to be identical, as to produce similar probability

Considering I.I.D, can we do a better probability estimation?

Random Variables

• All sampled random variables should be independent from one another
• Each sampling procedure have to be identical, as to produce similar probability

Considering I.I.D, can we do a better probability estimation?

• If they are I.I.D, we can approximate the probability using:
  • Probability Mass Function (discrete variable)
  • Probability Density Function (continuous variable)

Binomial Distribution

Binomial Distribution

$$\begin{array}{cc}f\left(x\right)& =(\frac{n}{x})p^x(1-p{)}^{n-x}& (\frac{n}{x})& =\frac{n!}{x!(n-x)!}\\ \text{(1)}\end{array}$$

Or simply denoted as: $X\sim B(n,p)$

Binomial Distribution

$$\begin{array}{cc}f\left(x\right)& =(\frac{n}{x})p^x(1-p{)}^{n-x}& (\frac{n}{x})& =\frac{n!}{x!(n-x)!}\\ \text{(1)}\end{array}$$

Or simply denoted as: $X\sim B(n,p)$

$$\begin{array}{cc}\mu & =n\cdot p\\ \sigma & =\sqrt{\mu \cdot (1-p)}\end{array}$$

Binomial Distribution

Geometric Distribution

Geometric Distribution

$$f\left(n\right)=P(X=n)=p(1-p{)}^{n-1},with:$$

$n$: Number of trials to get an event
$p$: The probability of getting an event
Or simply denoted as $X\sim G\left(p\right)$

Geometric Distribution

$$f\left(n\right)=P(X=n)=p(1-p{)}^{n-1},with:$$

$n$: Number of trials to get an event
$p$: The probability of getting an event
Or simply denoted as $X\sim G\left(p\right)$

$$\begin{array}{cc}\mu & =\frac{1}{p}\\ \sigma & =\sqrt{\frac{1-p}{p^2}}\end{array}$$

Poisson Distribution

Poisson Distribution

$$f\left(x\right)=\frac{e^{-\lambda }{\lambda }^x}{x!},with:$$

$x$: The number of expected events
$e$: Euler's number
$\lambda$: Average number of events in one time frame

Or simply denoted as $X\sim P\left(\lambda \right)$

$$\begin{array}{cc}\mu & =\lambda \\ \sigma & =\sqrt{\lambda }\end{array}$$

Uniform Distribution

Uniform Distribution

$$f\left(x\right)=\frac{1}{b-a}$$

Or simply denoted as $X\sim U(a,b)$

Uniform Distribution

$$f\left(x\right)=\frac{1}{b-a}$$

Or simply denoted as $X\sim U(a,b)$

$$\begin{array}{cc}\mu & =\frac{b+a}{2}\\ \sigma & =\frac{(b-a{)}^2}{12}\end{array}$$

Exponential Distribution

Exponential Distribution

$$f\left(x\right)=\lambda e^{-x\lambda },with:$$

$x$: Time needed to observe an event
$\lambda$: The rate for a certain event

Or simply denoted as $X\sim Exponential\left(\lambda \right)$

$$\mu =\sigma =\frac{1}{\lambda }$$

Gamma Distribution

• Exponential distribution is a gamma distribution without a shape parameter
• Essentially, gamma distribution finds its uses in similar cases as exponential distribution
• Relies on the gamma function $\Gamma \left(\alpha \right)$

Gamma Distribution

• Exponential distribution is a gamma distribution without a shape parameter
• Essentially, gamma distribution finds its uses in similar cases as exponential distribution
• Relies on the gamma function $\Gamma \left(\alpha \right)$

$$\begin{array}{cc}f\left(x\right)& =\frac{{\beta }^{\alpha }}{\Gamma \left(\alpha \right)}{x}^{\alpha -1}{e}^{-x\beta }\\ \Gamma \left(\alpha \right)& ={\int }_0^{\infty }{y}^{\alpha -1}{e}^{-y}dy,with:\end{array}$$

$\beta$: Rate ( $\lambda$ in exponential PDF)
$\alpha$: Shape
$\Gamma$: Gamma function
$e$: Euler number

Or simply denoted as $X\sim \Gamma (\alpha ,\beta )$

Gamma Distribution

• Exponential distribution is a gamma distribution without a shape parameter
• Essentially, gamma distribution finds its uses in similar cases as exponential distribution
• Relies on the gamma function $\Gamma \left(\alpha \right)$

$$\begin{array}{cc}f\left(x\right)& =\frac{{\beta }^{\alpha }}{\Gamma \left(\alpha \right)}{x}^{\alpha -1}{e}^{-x\beta }\\ \Gamma \left(\alpha \right)& ={\int }_0^{\infty }{y}^{\alpha -1}{e}^{-y}dy,with:\end{array}$$

$\beta$: Rate ( $\lambda$ in exponential PDF)
$\alpha$: Shape
$\Gamma$: Gamma function
$e$: Euler number

Or simply denoted as $X\sim \Gamma (\alpha ,\beta )$

If we were to assign the shape parameter $\alpha =1$, we get an exponential PDF.

Gamma Distribution

• Exponential distribution is a gamma distribution without a shape parameter
• Essentially, gamma distribution finds its uses in similar cases as exponential distribution
• Relies on the gamma function $\Gamma \left(\alpha \right)$

$$\begin{array}{cc}f\left(x\right)& =\frac{{\beta }^{\alpha }}{\Gamma \left(\alpha \right)}{x}^{\alpha -1}{e}^{-x\beta }\\ \Gamma \left(\alpha \right)& ={\int }_0^{\infty }{y}^{\alpha -1}{e}^{-y}dy,with:\end{array}$$

$\beta$: Rate ( $\lambda$ in exponential PDF)
$\alpha$: Shape
$\Gamma$: Gamma function
$e$: Euler number

Or simply denoted as $X\sim \Gamma (\alpha ,\beta )$

If we were to assign the shape parameter $\alpha =1$, we get an exponential PDF. Therefore, $Exponential\left(\lambda \right)\sim \Gamma (1,\lambda )$.

Gamma Distribution

• Exponential distribution is a gamma distribution without a shape parameter
• Essentially, gamma distribution finds its uses in similar cases as exponential distribution
• Relies on the gamma function $\Gamma \left(\alpha \right)$

$$\begin{array}{cc}f\left(x\right)& =\frac{{\beta }^{\alpha }}{\Gamma \left(\alpha \right)}{x}^{\alpha -1}{e}^{-x\beta }\\ \Gamma \left(\alpha \right)& ={\int }_0^{\infty }{y}^{\alpha -1}{e}^{-y}dy,with:\end{array}$$

$\beta$: Rate ( $\lambda$ in exponential PDF)
$\alpha$: Shape
$\Gamma$: Gamma function
$e$: Euler number

Or simply denoted as $X\sim \Gamma (\alpha ,\beta )$

If we were to assign the shape parameter $\alpha =1$, we get an exponential PDF. Therefore, $Exponential\left(\lambda \right)\sim \Gamma (1,\lambda )$.

$$\begin{array}{cc}\mu & =\frac{\alpha }{\beta }\\ \sigma & =\frac{\sqrt{\alpha }}{\beta }\end{array}$$

${\chi }^2$ Distributions

${\chi }^2$ Distributions

$$f\left(x\right)=\frac{1}{\Gamma \left(k/2\right)2^{k/2}}{x}^{k/2-1}{e}^{-x/2},with:$$

$k$: Degree of freedom
The rest are Gamma PDF derivations

Or simply denoted as $X\sim {\chi }^2\left(k\right)$

$$\begin{array}{cc}\mu & =k\\ \sigma & =\sqrt{2k}\end{array}$$

${\chi }^2$ Distributions

$$f\left(x\right)=\frac{1}{\Gamma \left(k/2\right)2^{k/2}}{x}^{k/2-1}{e}^{-x/2},with:$$

$k$: Degree of freedom
The rest are Gamma PDF derivations

Or simply denoted as $X\sim {\chi }^2\left(k\right)$

$$\begin{array}{cc}\mu & =k\\ \sigma & =\sqrt{2k}\end{array}$$

Relation to normal distribution?

Normal Distribution

Normal Distribution

$$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}exp\{-\frac{1}{2}(\frac{x-\mu }{\sigma }{)}^2\},with:$$

$x\in R:-\infty <x<\infty$
$\mu \in R:-\infty <\mu <\infty$
$\sigma \in R:0<\sigma <\infty$

Or simply denoted as $X\sim N(\mu ,\sigma )$

Normal Distribution

Normal Distribution

Normal Distribution

Binomial Test

Binomial Test

Remember we previously tossed a coin 10 times?

Binomial Test

Remember we previously tossed a coin 10 times?

set.seed(1)
S <- sample(c("H", "T"), 10, replace=TRUE, prob=rep(1/2, 2)) %T>% print()

##  [1] "T" "T" "H" "H" "T" "H" "H" "H" "H" "T"

length(E) / length(S)

## [1] 0.6

Binomial Test

Remember we previously tossed a coin 10 times?

set.seed(1)
S <- sample(c("H", "T"), 10, replace=TRUE, prob=rep(1/2, 2)) %T>% print()

##  [1] "T" "T" "H" "H" "T" "H" "H" "H" "H" "T"

length(E) / length(S)

## [1] 0.6

If it represents a Bernoulli trial, it should satisfy $P(X=6)$ in such a way that we cannot reject the $H_0$ when calculating its probability:

$$P(X=6)=(\frac{10}{6}){0.5}^6(1-0.5{)}^4$$

Binomial Test

Luckily, we do not need to compute it by hand

Binomial Test

Luckily, we do not need to compute it by hand (yet)

binom.test(x=6, n=10, p=0.5)

## 
##     Exact binomial test
## 
## data:  6 and 10
## number of successes = 6, number of trials = 10, p-value = 0.8
## alternative hypothesis: true probability of success is not equal to 0.5
## 95 percent confidence interval:
##  0.2624 0.8784
## sample estimates:
## probability of success 
##                    0.6

Kolmogorov-Smirnov Test

Kolmogorov-Smirnov Test

Let $X\sim Exponential\left(2\right):n=100$

set.seed(1); X <- rexp(n=100, rate=2)

Kolmogorov-Smirnov Test

Let $X\sim Exponential\left(2\right):n=100$

set.seed(1); X <- rexp(n=100, rate=2)

By imputing $\lambda$ variable, Kolmogorov-Smirnov can compute its goodness of fit

ks.result <- ks.test(X, pexp, rate=2)

Visual Examination

Visual Examination

Visual Examination

Visual Examination

Hey, that's a good start!

Visual Examination

Hey, that's a good start! This does not clearly suggest a specific distribution though :(

Visual Examination

Visual Examination

Quantile-Quantile Plot (QQ Plot) can give a better visual cue :)

Demonstration

Let $X\sim N(0,1):n=100$

set.seed(1)
X <- rnorm(n=100, mean=0, sd=1)

Demonstration

Let $X\sim N(0,1):n=100$

set.seed(1)
X <- rnorm(n=100, mean=0, sd=1)

Shapiro-Wilk

shapiro.test(X)

## 
##     Shapiro-Wilk normality test
## 
## data:  X
## W = 1, p-value = 1

Demonstration

Let $X\sim N(0,1):n=100$

set.seed(1)
X <- rnorm(n=100, mean=0, sd=1)

Anderson-Darling

nortest::ad.test(X)

## 
##     Anderson-Darling normality test
## 
## data:  X
## A = 0.16, p-value = 0.9

${\chi }^2$ and Normal Distribution

${\chi }^2$ and Normal Distribution

For demonstration purposes, we will re-use $X\sim N(0,1):n=100$

set.seed(1)
X <- rnorm(n=100, mean=0, sd=1)

We previously tested $X$ against Shapiro-Wilk and Anderson-Darling tests to indicate normality.

${\chi }^2$ and Normal Distribution

For demonstration purposes, we will re-use $X\sim N(0,1):n=100$

set.seed(1)
X <- rnorm(n=100, mean=0, sd=1)

We previously tested $X$ against Shapiro-Wilk and Anderson-Darling tests to indicate normality.

Now, we will raise it to the power of two

X2 <- X^2

${\chi }^2$ and Normal Distribution

Does it still follow a normal distribution?

shapiro.test(X2)

## 
##     Shapiro-Wilk normality test
## 
## data:  X2
## W = 0.7, p-value = 5e-13

Visual Examination

${\chi }^2$ and Normal Distribution

It does not follow normal distribution at all.

${\chi }^2$ and Normal Distribution

It does not follow normal distribution at all. Does it follow the ${\chi }^2$ distribution though?

${\chi }^2$ and Normal Distribution

It does not follow normal distribution at all. Does it follow the ${\chi }^2$ distribution though?

ks.test(X2, pchisq, df=1)

## 
##     One-sample Kolmogorov-Smirnov test
## 
## data:  X2
## D = 0.1, p-value = 0.2
## alternative hypothesis: two-sided

Visual Examination

Central Limit Theorem

Central Limit Theorem

Central Limit Theorem

Central Limit Theorem

How do we determine $n$?

Central Limit Theorem

How do we determine $n$?$\to$ Simulation

Central Limit Theorem

How do we determine $n$?$\to$ Simulation

1. Choose any distribution

Central Limit Theorem

How do we determine $n$?$\to$ Simulation

1. Choose any distribution
2. Generate $n$ random numbers using specified parameters

Central Limit Theorem

How do we determine $n$?$\to$ Simulation

1. Choose any distribution
2. Generate $n$ random numbers using specified parameters
3. Compute the mean and variance based on previous parameters $\to$ Use it to generate a normal distribution

Central Limit Theorem

How do we determine $n$?$\to$ Simulation

1. Choose any distribution
2. Generate $n$ random numbers using specified parameters
3. Compute the mean and variance based on previous parameters $\to$ Use it to generate a normal distribution
4. Reuse the parameters to re-iterate step 2

Central Limit Theorem

How do we determine $n$?$\to$ Simulation

1. Choose any distribution
2. Generate $n$ random numbers using specified parameters
3. Compute the mean and variance based on previous parameters $\to$ Use it to generate a normal distribution
4. Reuse the parameters to re-iterate step 2
5. Conduct the simulation for an arbitrary number of times (e.g. for convenience, 1000)

Central Limit Theorem

How do we determine $n$?$\to$ Simulation

1. Choose any distribution
2. Generate $n$ random numbers using specified parameters
3. Compute the mean and variance based on previous parameters $\to$ Use it to generate a normal distribution
4. Reuse the parameters to re-iterate step 2
5. Conduct the simulation for an arbitrary number of times (e.g. for convenience, 1000)
6. Calculate mean from all generated data $\to$ Make a histogram and compare it with step 3

Central Limit Theorem

How do we determine $n$?$\to$ Simulation

1. Choose any distribution
2. Generate $n$ random numbers using specified parameters
3. Compute the mean and variance based on previous parameters $\to$ Use it to generate a normal distribution
4. Reuse the parameters to re-iterate step 2
5. Conduct the simulation for an arbitrary number of times (e.g. for convenience, 1000)
6. Calculate mean from all generated data $\to$ Make a histogram and compare it with step 3
7. Does not fit normal distribution? $\to$ Increase $n$

Central Limit Theorem

Why should you care?